# Miscellaneous Mathematics (open thread)



## opaltiger (Sep 23, 2009)

Because maths is awesome! Plus I want to practice a little. And I have a nice problem!

Let two vertices of a rectangle lie on the x-axis and the other two on the curve y = e^(-x^2). a) Find an expression for the area of the rectangle. b) Find the maximum area.

So we know that the curve will be symmetrical over the y-axis, because the function is even (since -x^2 = x^2). Thus we can determine that the vertices on the curve will have the coordinates (x,y) and (-x,y). This means that one side (the base) of the rectangle will be 2x (x - (-x)). The other side (the height) is y, or e^(-x^2). Thus we can express the area as

A = 2x * e^(-x^2)

If we then differentiate:

dy/dx = 2x * e^(-x^2) * (-2x) + e^(-x^2) * 2
dy/dx = e^(-x^2)(-4x^2 + 2)

equal to zero:

e^(-x^2)(-4x^2 + 2) = 0
-4x^2 + 2 = 0 (because e^n where n is any real number is always positive)
2 = 4x^2
1/2 = x^2
x = +/- 1/sqrt(2)

substitute back into the area expression (taking the positive result, because area cannot be negative):

A(max) = 2x * e^(-x^2)
A(max) = 2/sqrt(2) * e^(-1/sqrt(2))^2
A(max) = sqrt(2) * e^(1/2)

And this is the kind of maths I really like, both because it is such a pleasure to work your way through a longer problem and get the right answer at the end and because you can actually see what this kind of thing might be used for.

So. This is an open thread; feel free to ask for help, post your own interesting problems, or whatever.


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## Tarvos (Sep 23, 2009)

i'll contribute to this thread if we have a need for some advanced algebra stuff as i'm of course studying to be an engineer

and i've done diff. eqs and all that


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## octobr (Sep 23, 2009)

lol maybe you can explain shit right when my calculus class cannot. 

god I love math but what can I do if I am not given the challenges I need.


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## Aobaru (Sep 24, 2009)

I'm taking high school-level Physics and Algebra 2, so I doubt I could be of any help here...


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## Cubaser (Sep 24, 2009)

I'm only taking Geometry, so you kinda got me a little lost there. But I'd like to help with anything else!


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## Tarvos (Sep 25, 2009)

yeah i've done uni level calculus go


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## 1. Luftballon (Sep 27, 2009)

opaltiger said:


> A(max) = sqrt(2) * e^(1/2)


can't this be simplified to A(max) = sqrt(2) * sqrt(e) = sqrt(2e) ?


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## opaltiger (Sep 27, 2009)

Fair point. I had the answer listed as the former, but yes.


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## octobr (Sep 30, 2009)

Walk me through sum derivative shit, yo.

They're asking to find dy/dx of y=(x^2 + 3)/x with a. quotient rule and b. I keep coming up with different things and gah.


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## Butterfree (Sep 30, 2009)

Well, with the quotient rule, we would set f(x) as the numerator, x^2 + 3, and g(x) as the denominator, x. Then:

y' = (f'(x)g(x) - f(x)g'(x))/(g(x))^2 = (2x*x - (x^2 + 3)*1)/x^2

y' = (2x^2 - x^2 - 3)/x^2

y' = (x^2 - 3)/x^2 = 1 - 3/x^2

Though I'm not sure exactly what you mean; are you supposed to be finding it by some other means as well? If you're supposed to do it by any other method you feel like, the most straightforward one would be to just rewrite it to y = x^2/x + 3/x = x + 3/x and then differentiate each part separately to get y' = 1 + (-3/x^2) = 1 - 3/x^2.


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## octobr (Sep 30, 2009)

That's what I thought, ok, thanks. 

Oh sorry I obviously forgot to type out what b was. They want me to divide out first and then differentiate? Cept like ... the only way to really divide it out is just far too complicated, bah. I can't imagine why you would.


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## Dewgong (Oct 1, 2009)

i wish i was good at math. i might come here if i need help with homework or something...


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## Aobaru (Oct 3, 2009)

Does anyone else here think Physics is _extremely_ fascinating?


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## Skroy (Oct 3, 2009)

Cool, a math thread. 8D Maybe I can ask for help with my homework whenever I'm in a rut....

I got a two basic questions I want to ask because I see them being used here which has got me curious, and plus I would like to learn them because I have no knowledge of them... yet:

1) What is "dy/dx"? (This is my first time seeing this just so you know). 

2) What is the quotient rule? (Also my first time).

EDIT: Actually, I got another one: "3) What is sum derivative?" (I'm pretty sure I'm gonna learn this next class after the test).
As you can see, I'm really curious to know what these are. :3


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## Butterfree (Oct 3, 2009)

dy/dx is Leibniz's notation for a derivative, which is basically a function describing a base function's rate of change. The simplest way to explain it is that if you for instance plot a graph with time on the x axis and location on the y axis, you can create a function describing how your location changes with time as you, say, drive in a straight line. The derivative of this function will describe how your speed changes during this journey. (Then you could differentiate (i.e. find the derivative of) _that_, and you'd get a function describing how your acceleration changes.)

The quotient rule is a rule of differentiation which states that the derivative of a fraction, if f(x) is the numerator, g(x) is the denominator and f'(x) and g'(x) are their respective derivatives, is (f'(x)*g(x) - f(x)* g'(x))/(g(x))^2.


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## octobr (Oct 3, 2009)

In other words, for the quotient rule (this is how I think of it, it's the same but is just much less messy):

if you wanna take the derivative of u/v, where u and v are two functions.

(v*du - u*dv)/v^2

I just happen to hate function notations. So many parentheses and blah.


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## Skroy (Oct 3, 2009)

Thanks for explaining what those two are guys. ^w^ I would like to clarify one thing though: _dy/dx_ is the same as _f'(x)_ right?


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## octobr (Oct 4, 2009)

Yep. There's like twenty bagrillion different ways to write out notation for derivatives.


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## opaltiger (Oct 4, 2009)

I like y' because I am lazy but dy/dx looks cooler.


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## Lorem Ipsum (Oct 4, 2009)

Isn’t dy/dx the function for finding the gradient of a line?


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## Butterfree (Oct 4, 2009)

...which is the same as the derivative.


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## Tarvos (Oct 5, 2009)

i use dy/dx more because i keep solving differential equations and having to separate variables


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## Skroy (Oct 25, 2009)

Ugh, I hate to admit this but....

"If d/dx (f(4x^5)) = 9x^4, calculate f'(x)"

Surprisingly, I'm at a lost for this question. =S Um, I'm not sure of what to do.... Chain Rule perhaps?

EDIT: Oh wait, never mind, I just got the answer. Heh, silly me. ^^; For those interested:

Firstly, I'll let F be equal to the derivative of f(4x^5) [Just look at the question].
F = d/dx ((f(4x^5))
F = f(4x^5) * 20x^4 [<- By chain rule]
9x^4 = f(4x^5) * 20x^4
9x^4/20x^4 = f(4x^5)

Ans: f'(x) = 9/20


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## Tarvos (Oct 26, 2009)

I tie a rope around the earth. This rope fits completely perfectly around the earth (i.e. there is no distance between rope and surface and the rope is completely stretched so that it is not too long or too short).

But Tailsy has noticed this, sneaky girl. To be a bitch, and to undermine my antics, she stealthily cuts the rope at night, and adds an extra meter, and reties the rope. During the same night, my ant colony escapes, which were bound by the perimeter of my rope. Can my ant colony escape now that Tailsy has changed the rope? Assume all my ants walk in a single file line and don't climb over each other to cover the rope.


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## Peter Shadeslayer (Nov 8, 2009)

*Just Wondering*

^^"

This isn't for homework or anything. I just don't remember how to do this and no one is on MSN or anything. 

f(x) = 3x^3 + 4x + 3

I took Algebra last year and now I'm getting rusty because I'm in Geometry now and we don't do a lot like this. ;)


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## spaekle (Nov 8, 2009)

*Re: Just Wondering*

Oh hey, we just finished doing that.

I think you need to specify what you're supposed to find for x. Like, if it were f(4), it would be

f(4) = 3(4)^3 + 4(4) + 3
*(Spaekle is dumb, see below)*
= 12^3 + 16 + 3
= 1728 + 16 + 3
= 1747 

so f(4) = 1747 unless I fucked up again. I sort of suck at math, as a disclaimer. Feel free to yell at me for my mathematical insolence if I'm doing it wrong.


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## Peter Shadeslayer (Nov 8, 2009)

*Re: Just Wondering*

Actually I think that the solution would be

f(4) = 3(64) + 16 + 3
= 192 + 16 + 3
= 211

although I'm not sure.

But I mean just to simplify that equation as much as possible. :)


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## spaekle (Nov 8, 2009)

*Re: Just Wondering*

It might be. I felt like there was an order of operations fail in there somewhere. :(

Edit: Oh yeah, exponent.
Edit 2: Calculator gives me your answer and calculator is always right. This is what happens when I try it without a calculator. :v;;;


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## opaltiger (Nov 8, 2009)

*Re: Just Wondering*

Well, it kinda depends what you need to do with the function. If the point is to find roots, then you'd equal it to 0 and solve for x, although I'm pretty sure there's a pair of complex roots. Otherwise there are a whole bunch of questions you could ask about the function, so...


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## Tarvos (Nov 9, 2009)

Read out the whole question; just say SOLVE DA FUNCTION doesn't really work


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## Butterfree (Nov 10, 2009)

Watershed said:


> I tie a rope around the earth. This rope fits completely perfectly around the earth (i.e. there is no distance between rope and surface and the rope is completely stretched so that it is not too long or too short).
> 
> But Tailsy has noticed this, sneaky girl. To be a bitch, and to undermine my antics, she stealthily cuts the rope at night, and adds an extra meter, and reties the rope. During the same night, my ant colony escapes, which were bound by the perimeter of my rope. Can my ant colony escape now that Tailsy has changed the rope? Assume all my ants walk in a single file line and don't climb over each other to cover the rope.


I don't get it. Can they escape from where to where? How exactly is the rope retied?


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## Tarvos (Nov 25, 2009)

The ant colony needs to cross under the rope to escape. The rope is retied exactly as it was, it's just a metre longer.


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## Zhorken (Nov 26, 2009)

Watershed said:


> During the same night, my ant colony escapes, which were bound by the perimeter of my rope. Can my ant colony escape now that Tailsy has changed the rope?


Well they just did. ...


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## Tarvos (Nov 26, 2009)

I mean can they escape under the rope because I can catch them if they don't. Smartass.


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## Jack_the_White (Dec 10, 2009)




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